Coordinate Geometry

**Punch** · July 11th 2010, 07:40 AM

In the diagram above, triangle $ACD$ is such that the coordinates of $A$ and $B$ are $(2, 12)$ and $(6, 9)$ respectively. $BE$ is parallel to $CD$ and $BC = \frac{2}{3}AC$ . The gradient of $BE$ is $\frac{2}{7}$ . The equation of $AD$ is $3y=5x+26$ .

Calculate the coordinates of C.

**red_dog** · July 11th 2010, 08:39 AM

Let $M_1(x_1,y_1), \ M_2(x_2,y_2)$ be two points and $M(x_M,y_M)$ a point on the segment $M_1M_2$ such as $\displaystyle\frac{M_1M}{MM_2}=k$ .

Then $x_M=\displaystyle\frac{x_1+kx_2}{1+k}, \ y_M=\displaystyle\frac{y_1+ky_2}{1+k}$

In this case $\displaystyle\frac{AB}{AC}=\frac{1}{2}$

Then $x_B=\displaystyle\frac{x_A+\frac{1}{2}x_C}{1+\frac {1}{2}}$

$y_B=\displaystyle\frac{y_A+\frac{1}{2}y_C}{1+\frac {1}{2}}$

You know the coordinates of A and B, so you can find the coordinates of C.

**Soroban** · July 11th 2010, 08:43 AM

Hello, Punch!

I assume there are more parts to this problem.
There is way too much information . . .

In the diagram, $\Delta ACD$ has points: $A(2,12)$ and $B(6,9).$

$BE$ is parallel to $CD$ and $BC = \frac{2}{3}AC$ .
The gradient of $BE$ is $\frac{2}{7}$ .
The equation of $AD$ is: $3y\:=\:5x+26$ .

Calculate the coordinates of $C.$

Here is a simplified diagram:

Code:

      |
      |       A
      | (2,12)o - - +
      |        \    :
      |         \   :
      |          \  :
      |           \ :
      |            \:
      |       B(6,9)o - - - - +
      |              \        :
      |               \       :
      |                \      :
      |                 \     :
      |                  \    :
      |                   \   :
      |                    \  :
      |                     \ :
      |                      \:
      |                     C o
      |
  - - + - - - - - - - - - - - - - -
      |

$\text{Since }BC = \frac{2}{3}AC,\; BC\text{ is twice }AB$

Going from $A$ to $B$ , we move: 4 units right and 3 units down.

So, going from $B$ to $C$ , we move: 8 units right and 6 units down.

Therefore, point $C$ is $(14,3).$

**Punch** · July 11th 2010, 04:26 PM

you are right soroban

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